已知正项数列{an},a1^3+a2^3+a3^3++an^3=sn^2已知正项数列{an},a1^3+a2^3+a3^3++an^3=sn^2求{an}的通项设bn=3^n+(-1)^(n-1*入*2^an)(入为非零整数),求入使得对任意的正整数都有b(n+1)>bn
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/02 00:04:13
![已知正项数列{an},a1^3+a2^3+a3^3++an^3=sn^2已知正项数列{an},a1^3+a2^3+a3^3++an^3=sn^2求{an}的通项设bn=3^n+(-1)^(n-1*入*2^an)(入为非零整数),求入使得对任意的正整数都有b(n+1)>bn](/uploads/image/z/10322244-36-4.jpg?t=%E5%B7%B2%E7%9F%A5%E6%AD%A3%E9%A1%B9%E6%95%B0%E5%88%97%7Ban%7D%2Ca1%5E3%2Ba2%5E3%2Ba3%5E3%2B%2Ban%5E3%3Dsn%5E2%E5%B7%B2%E7%9F%A5%E6%AD%A3%E9%A1%B9%E6%95%B0%E5%88%97%7Ban%7D%EF%BC%8Ca1%5E3%2Ba2%5E3%2Ba3%5E3%2B%2Ban%5E3%3Dsn%5E2%E6%B1%82%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E8%AE%BEbn%3D3%5En%2B%28-1%29%5E%28n-1%2A%E5%85%A5%2A2%5Ean%29%EF%BC%88%E5%85%A5%E4%B8%BA%E9%9D%9E%E9%9B%B6%E6%95%B4%E6%95%B0%EF%BC%89%EF%BC%8C%E6%B1%82%E5%85%A5%E4%BD%BF%E5%BE%97%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84%E6%AD%A3%E6%95%B4%E6%95%B0%E9%83%BD%E6%9C%89b%28n%2B1%29%3Ebn)
已知正项数列{an},a1^3+a2^3+a3^3++an^3=sn^2已知正项数列{an},a1^3+a2^3+a3^3++an^3=sn^2求{an}的通项设bn=3^n+(-1)^(n-1*入*2^an)(入为非零整数),求入使得对任意的正整数都有b(n+1)>bn
已知正项数列{an},a1^3+a2^3+a3^3++an^3=sn^2
已知正项数列{an},a1^3+a2^3+a3^3++an^3=sn^2
求{an}的通项
设bn=3^n+(-1)^(n-1*入*2^an)(入为非零整数),求入使得对任意的正整数都有b(n+1)>bn
已知正项数列{an},a1^3+a2^3+a3^3++an^3=sn^2已知正项数列{an},a1^3+a2^3+a3^3++an^3=sn^2求{an}的通项设bn=3^n+(-1)^(n-1*入*2^an)(入为非零整数),求入使得对任意的正整数都有b(n+1)>bn
Sn^2=a1^3+a2^3+...+an^3
S(n-1)^2=a1^3+a2^3+...+a(n-1)^3
相减有
(Sn-S(n-1))(Sn+S(n-1)=An^3
Sn+S(n-1)=An^2
Sn+Sn-An=An^2
2Sn=An^2+An
2S(n-1)=A(n-1)^2+A(n-1)
相减有
2An=An^2-A(n-)^2+An-A(n-1)
(An+A(n-1))*(An-A(n-1)-1)=0
An为正项数列,An+A(n-1)>0
所以
An-A(n-1)=1
所以An为等差数列,d=1
A1^3=S1^2=A1^2
A1=1
1)
An=A1+(n-1)D=1+(n-1)=n
2)
n-1*入*2^an看不懂,为什么出来一个1